Question: Simplify and expand the following expression: $ \dfrac{5x + 1}{4x + 1}+\dfrac{4x - 8}{x + 3} $
Solution: In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(4x + 1)(x + 3)$ Multiply the first term by $\dfrac{x + 3}{x + 3}$ $ \begin{align*} \dfrac{5x + 1}{4x + 1} \times \dfrac{x + 3}{x + 3} & = \dfrac{(5x + 1)(x + 3)}{(4x + 1)(x + 3)} \\ & = \dfrac{5x^2 + 16x + 3}{(4x + 1)(x + 3)}\end{align*} $ Multiply the second term by $\dfrac{4x + 1}{4x + 1}$ $ \begin{align*} \dfrac{4x - 8}{x + 3} \times \dfrac{4x + 1}{4x + 1} & = \dfrac{(4x - 8)(4x + 1)}{(x + 3)(4x + 1)} \\ & = \dfrac{16x^2 - 28x - 8}{(x + 3)(4x + 1)}\end{align*} $ Now we have: $ = \dfrac{5x^2 + 16x + 3}{(4x + 1)(x + 3)} + \dfrac{16x^2 - 28x - 8}{(x + 3)(4x + 1)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{5x^2 + 16x + 3 + 16x^2 - 28x - 8}{(4x + 1)(x + 3)} $ $ = \dfrac{21x^2 - 12x - 5}{(4x + 1)(x + 3)}$ Expand the denominator: $ = \dfrac{21x^2 - 12x - 5}{4x^2 + 13x + 3}$